The work done in shifting a particle of mass m from the centre of the earth to the surface of the earth is (Where M is the mass of the earth and R is the radius of the earth)

Question

The work done in shifting a particle of mass m from the centre of the earth to the surface of the earth is (Where M is the mass of the earth and R is the radius of the earth) (A) - (GMm/R) (B) + (GMm/R) (C) + (GMm/2R) (D) - (GMm/2R)

Tardigrade Admin · Accepted Answer

Gravitational potential energy at the centre of the earth is

u_{i} = - \frac{3 GM m}{2 R}

Gravitational potential energy at the surface of the earth is

u_{f} = - \frac{GM m}{R}

Work done,

W = U_{f} - U_{i}

= - \frac{GM m}{R} - (- \frac{3 G m}{2 R})

W = - \frac{GM m}{R} + \frac{3 GM m}{2 R}

= + \frac{GM m}{2 R}

Q. The work done in shifting a particle of mass $m$ from the centre of the earth to the surface of the earth is (Where $M$ is the mass of the earth and $R$ is the radius of the earth)

$- \frac{GM m}{R}$

$+ \frac{GM m}{R}$

$+ \frac{GM m}{2 R}$

$- \frac{GM m}{2 R}$

Q. The work done in shifting a particle of mass m from the centre of the earth to the surface of the earth is (Where M is the mass of the earth and R is the radius of the earth)

−RGMm​

+RGMm​

+2RGMm​

−2RGMm​

Gravitational potential energy at the centre of the earth is ui​=−2R3GMm​ Gravitational potential energy at the surface of the earth is uf​=−RGMm​ Work done, W=Uf​−Ui​ =−RGMm​−(−2R3Gm​) W=−RGMm​+2R3GMm​ =+2RGMm​

Q. The work done in shifting a particle of mass $m$ from the centre of the earth to the surface of the earth is (Where $M$ is the mass of the earth and $R$ is the radius of the earth)

$- \frac{GM m}{R}$

$+ \frac{GM m}{R}$

$+ \frac{GM m}{2 R}$

$- \frac{GM m}{2 R}$